Taking the Partial Derivative of a Function

Date: 09/06/2002 at 04:16:02
From: Robert Curl
Subject: Partial derivative
Dear Dr. Math,
Given an equation, for example, x+y=0, is it true that if we perform
the same operation on both sides we could have the same results on
both sides?
For example, if we add 6 to both sides, and on the left side we have
(x+y)+6, and on the right side we have 0+6=6, we know that the results
on the left and the right sides are equal, that is, (x+y)+6=0+6.
But how about the operation of partial derivative? For the equation
x+y=0, if we perform the operation of partial derivative on the left
side and let F(x,y)=x+y, we have (partial F)/(partial x)=1 and
(partial F)/(partial y)=1, but on the right, we know the result is
zero, which is not equal to the result on the left. One must insert
the solution, y=y(x)=-x, into the equation; that is, on the left side,
F(x,y)=x+y, x=x, y=-x, then dF/dx=0, which is equal to the result
obtained on the right.
From the above, I think the left and right sides of an equation are
not always symmetric. For the operations plus, minus, etc., we could
always have same results on both sides after performing the same
operation on both sides, but for the operation of derivative, the
results on both sides are not always equal.
Am I right? I need your help. Thank you very much.
Sincerely,
Robert Curl

Date: 09/06/2002 at 08:46:01
From: Doctor Jerry
Subject: Re: Partial derivative
Hi Robert,
The partial derivative with respect to a variable is an operation
that is defined in terms of functions. That is, one takes the partial
derivative of a function.
If you apply the partial derivative operation to an equation, you are
assuming that the equation defines one of the variables in terms of
the others. Using x+y=0 as an example and assuming that we want to
differentiate with respect to x, we think of y as the function of x
that we would find if we could solve this equation for y in terms of
x. Differentiating with respect to x then gives (y_x means the
derivative of y with respect to x)
1 + y_x = 0
and so
y_x = -1
This is correct becasue y = -x and we may differentiate the function
f(x,y) = -x with respect to x and find f_x = -1. In this case we are
using y as a symbol for f.
- Doctor Jerry, The Math Forum
http://mathforum.org/dr.math/

Date: 09/06/2002 at 08:47:31
From: Doctor Mitteldorf
Subject: Re: Partial derivative
Dear Robert,
It is easy to get into trouble manipulating symbols according to
preset rules, without thinking about what those symbols mean. In this
case, the equation x+y=0 might mean that we are considering a function
F(x,y) defined on the (x,y) plane to be the sum of the x and y
coordinates. Perhaps we are interested in the locus of points on that
plane where the function F has the value 0. In this case, it might be
a meaningful operation to take the partial derivative of the function
F with respect to x; but it wouldn't make sense to differentiate the
right side, because the 0 on the right represents a value, not a
function.
Of course, we might have a different situation, and the same equation
x+y=0 might mean something different: perhaps y is tied to x in such a
way as to guarantee that x+y is always 0. Then the partial
differentiation with respect to x would itself not be a meaningful
operation, since y is inextricably tied to x, and x can never be
varied while y is held constant.
My perspective is that you start with an understanding of a situation
in the world, and model that in relations among numbers. Ultimately it
is our understanding of the world and of the way that our model
represents it that determines what mathematical manipulations are
permitted and appropriate.
I am a scientist and statistician - not a pure mathematician.
Mathematicians might have a different perspective; it is their
discipline to define precisely the circumstances under which certain
purely formal symbolic manipulations can be performed, with internal
consistency in the results. I would be interested to see what a pure
mathematician might say about the paradox which you have uncovered.
- Doctor Mitteldorf, The Math Forum
http://mathforum.org/dr.math/